First, an introductory question:
There are $3n$ couples. The $6n$ people are each randomly allocated to one of three rooms, so that each room has $2n$ people. What is the probability that every couple is separated?
I made a crude approximation for large $n$: $\left(\frac{2}{3}\right)^{3n}$, because (assuming for the sake of simplicity that each couple consists of a husband and a wife) for each husband, there is an approximately $\frac{2}{3}$ chance that his wife is in a different room.
My question is:
It turns out that my approximation approaches exactly half of the exact value, as $n\to\infty$. Is there an intuitive explanation for this?
In case you're interested, the exact value is $\dfrac{\binom{3n}{2n}\binom{2n}{n}2^{3n}}{\binom{6n}{2n}\binom{4n}{2n}}$. Explanation:
- The denominator is the total number of ways to divide the $6n$ people
into three rooms. First, among the $6n$ people, we choose $2n$ people
to go to the first room, so $\binom{6n}{2n}$. Then among the
remaining $4n$ people we choose $2n$ people to go to the second room, so
$\binom{4n}{2n}$. Then the remaining $2n$ people go to the third
room. - The numerator is the total number of ways in which each couple is
separated. First, among the $3n$ couples, we choose $2n$ couples to
each be represented in the first room (one person from each couple),
so $\binom{3n}{2n}$. Then we allocate the $n$ remaining couples into
the second and third rooms (each room getting one member of each of
these couples), so the second room must now get an additional $n$
people, and these are chosen among the $2n$ people whose spouse went
to the first room, so $\binom{2n}{n}$. Then the remaining $n$ people
go to the third room. Then each couple has two ways of being separated among two rooms, so
$2^{3n}$.
It can be shown algebraically or by Wolfram that my approximation, $\left(\frac{2}{3}\right)^{3n}$, approaches exactly half of the exact value, as $n\to\infty$. I'm looking for an intuitive explanation.
(The introductory question was inspired by this question.)
Best Answer
I argue that there is no intuitive explanation, in the sense that the $2$ has a combinatorial background.
Stirling's approximation $n!\sim \sqrt{2\pi n}(n/e)^n$ and a very satisfying computation (since most terms cancel out) leaves you with twice your result. Most notably, the extra $2$ is the result of the leading square root term in Stirling's formula. Since this is a very low order contribution, it is rather a "relic" of taking the limit.
I do not argue that there is no interpretation. Terms like these usually have an impact on the variance of the limiting normal distribution in the CLT, which is then again the limit of the variances for the finite size laws. However, I don't think this was the (combinatorial) intuition you asked for.
Finally, here's an easier example to illustrate my point: the hypergeometric distribution, say $\binom{2n}{n}\binom{4n}{2n}/\binom{6n}{3n}\sim\sqrt{\frac{3}{2\pi}}$, i.e. the probability to get $n$ out for $2n$ hits when trying $3n$ out of $6n$ times.